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When 1 g of a cube of manganese (IV) oxide was added to 100 cm3 of hydrogen peroxide, water and oxygen were produced. Manganese oxide was not used up in the reaction. Based on this information, which of the following is likely to increase the rate formation of the products?

Using 2 grams of manganese oxide
Using 50 cm3 of hydrogen peroxide
Replacing the cube of manganese (IV) oxide with its powder
Removing manganese (IV) oxide from the reaction mixture

Respuesta :

znk

Answer:

[tex]\boxed{\text{Replacing the cube of manganese(IV) oxide with its powder}}[/tex]

Explanation:

The MnOā‚‚ is acting as a catalyst: it takes part in the reaction but can be recovered at the end.

A. Using a 2 g cube of MnOā‚‚

The surface area of the catalyst would double, so the rate would probably double. Technically, this option is correct, but I don't think it is the one the questioner intended.

B. Using 50 cmĀ³ of Hā‚‚Oā‚‚

No effect. The rate depends on how fast the Hā‚‚Oā‚‚ molecules can reach the surface of the catalyst, and the surface area hasn't changed.

C. Using powdered MnOā‚‚

This would increase the rate enormously, because the powdered catalyst has a much greater surface.

D. Removing MnOā‚‚

The rate would decrease to near-zero, because there is no catalyst.

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